The Lorentz group. Generate the group SO(3,1) To construct representa;ons a more convenient (non- Hermi;an) basis is. i j ijk k. j ijk k.

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Fermons S 1 2

The Lorentz group Rotatons J Boosts K [ J, J ] ε J [ J, K ] ε K [ K, K ] ε J j jk k j jk k j jk k } Generate the group SO(3,1) ( M ( x x ) J M K M ) 1 ρσ ρ σ x σ ρ x 2 ε jk jk 0 To construct representa;ons a more convenent (non- Herm;an) bass s 1 N ( ) 2 J + K [ N, N ] ε N [ N, N ] ε N j jk k [ N, N ] 0 j jk k j } SU (2) SU (2) representaton ( n, m)

The Lorentz group Rotatons J Boosts K [ J, J ] ε J [ J, K ] ε K [ K, K ] ε J j jk k j jk k j jk k } Generate the group SO(3,1) ( M ( x x ) J M K M ) 1 ρσ ρ σ x σ ρ x 2 ε jk jk 0 To construct representa;ons a more convenent (non- Herm;an) bass s 1 N ( ) 2 J + K [ N, N ] ε N [ N, N ] ε N j jk k [ N, N ] 0 j jk k j SU (2) L SU (2) R J N + N ( n, m) J n + m Representa;ons (0,0) scalar J0 ( 1 2,0), (0, 1 2 ) LH and RH "spnors" J 1 2 ( 1, 1 ) vector J1, etc 2 2

Weyl spnors (,0) (0, ) 1 1 2 2 L R 2- component spnors of SU(2) Rota;ons and Boosts L( R) SL( R) L( R) S L( R) e α 2.σ S L( R) e ν 2.σ : Rotatons : Boosts σ 1 0 1 1 0, σ 0 2 0, σ 1 0 3 0 1

Weyl spnors (,0) (0, ) 1 1 2 2 L R Rota;ons and Boosts Drac spnor Can combne S L( R) L( R) L( R) 2- component spnors of SU(2), L R to form a 4- component Drac spnor S L( R) e α.σ 2 : Rotatons S L( R) e ν.σ 2 : Boosts L R where γ 0 Lorentz transforma;ons ω 0 0 I I 0 e, ωσ ω σ γ, γ ω 0 σ, γ σ o, γ γ 0 γ 1 γ 2 γ 3 I 0 5 0 I ωσ µν µν µν 2 µ ν j boosts, ω rotatons, j 1,2,3 Weyl bass

Weyl spnors (,0) (0, ) 1 1 2 2 L R Rota;ons and Boosts L( R) L( R) L( R) Drac spnor Can combne S 2- component spnors of SU(2), L R to form a 4- component Drac spnor S L( R) e α.σ 2 : Rotatons S L( R) e ν.σ 2 : Boosts L R where γ 0 Lorentz transforma;ons γ µ (Drac gamma matrces new 4- vector ) { } 0 I I 0 e, ωσ ω σ γ, γ ω 0 σ, γ σ o, γ γ 0 γ 1 γ 2 γ 3 I 0 5 0 I µ ν µ ν ν µ µν γ, γ γ γ + γ γ 2g Clfford Algebra ωσ µν µν µν 2 µ ν

Weyl spnors (,0) (0, ) 1 1 2 2 L R Rota;ons and Boosts L( R) L( R) L( R) Drac spnor Can combne S 2- component spnors of SU(2), L R to form a 4- component Drac spnor S L( R) e α.σ 2 : Rotatons S L( R) e ν.σ 2 : Boosts L R where γ 0 Lorentz transforma;ons { } 0 I I 0 γ, γ γ γ + γ γ 2g µ ν µ ν ν µ µν e, ωσ ω σ γ, γ ω 0 σ, γ σ o, γ γ 0 γ 1 γ 2 γ 3 I 0 5 0 I γ µ (Drac gamma matrces new 4- vector ) ωσ µν µν µν 2 µ ν Note : L( R) 1 2 (1 γ 5 )

The Drac equa;on Fermons descrbed by 4- cpt Drac spnors L R γ 0 L R + R L Lorentz nvarant 2 2 3+1 S L( R) e α 2.σ S L( R) e ν 2.σ : Rotatons : Boosts

The Drac equa;on Fermons descrbed by 4- cpt Drac spnors L R γ 0 L R + R L Lorentz nvarant New 4- vector γ µ

The Drac equa;on Fermons descrbed by 4- cpt Drac spnors L R γ 0 L R + R L Lorentz nvarant New 4- vector γ µ Lorentz nvarant Lagrangan L - γ µ µ m g µν Dag(1, 1, 1, 1) ( g µν Srednck!) 2 2 3+1

The Drac equa;on Fermons descrbed by 4- cpt Drac spnors L R γ 0 L R + R L Lorentz nvarant New 4- vector γ µ 2 2 3+1 Lorentz nvarant Lagrangan L - γ µ µ m g µν Dag(1, 1, 1, 1) ( g µν Srednck!) From Euler Lagrange equa;on obtan the Drac equa;on ( γ µ µ m) 0 δs 0 L φ µ L ( µ φ) 0 Euler Lagrange equa;on

The Drac equa;on Fermons descrbed by 4- cpt Drac spnors γ 0 L R + R L Lorentz nvarant New 4- vector γ µ The Lagrangan L - γ µ µ m ( g µν Dag(1, 1, 1, 1) ) From Euler Lagrange equa;on obtan the Drac equa;on ( γ µ µ m) ( m) 0 p µ (m,0,0,0), ( γ 0 1) 1 1 1 1 0 2- components projected out

The Drac equa;on Fermons descrbed by 4- cpt Drac spnors γ 0 L R + R L Lorentz nvarant New 4- vector γ µ The Lagrangan L - γ µ µ m ( g µν Dag(1, 1, 1, 1) ) From Euler Lagrange equa;on obtan the Drac equa;on ( γ µ µ m) ( m) 0 p µ (m,0,0,0), ( γ 0 1) 1 1 1 1 ( γ µ µ + m) ( γ µ µ m) 0 ( 2 + m 2 ) 0 0 2- components projected out Momentum components off shell Projected out

The Drac equa;on Fermons descrbed by 4- cpt Drac spnors γ 0 L R + R L New 4- vector γ µ Lorentz nvarant The Lagrangan L - γ µ µ m γ µ µ γ 0 (γ 0 0 γ ) I 0 0 I I 0 0 I 0 σ 0 0 σ 0 0 I I 0 0 σ σ 0 σ µ L µ L + R σ µ µ R σ µ ( 1,σ ), σ µ ( 1, σ )

Can construct LH spnors out of RH an;spnors and vce- versa L σ 2 R * ( 1 2,0) R σ 2 L * (0, 1 2 ) Proof : L eσ!" 2.ν L?!" R e σ 2.ν R!" σ 2 * R σ 2 e σ *!" 2.ν" * R σ 2 e σ eσ!" * 2.ν" σ 2 σ 2 * 2 R.ν" * σ 2 R usng σ 2 σ * σ 2 σ

Majorana fermons Maj L R σ 2 L * only 2 ndependent components Majorana mass m( L R + R L ) M L t σ 2 L M! L L

Party x µ x x x x 0 0 (, ) (, ) J J K K N N N N N J + K 0 I L R γ 0 R L I 0.e. L σ µ L R σ µ R L Knetc L σ µ µ L + R σ µ µ R preserves party: 0 0, σ µ (1,σ ), σ µ (1, σ ) (neutrnos volate party).e. L σ µ µ L R σ µ µ R

Charge conjuga;on L Knetc L σ µ µ L + R σ µ µ R nvarant under L σ 2 R *, R σ 2 L * (σ 2 σ µ σ 2 σ 2 µt ) EM and QCD nterac;ons preserve charge conjuga;on: L EM Q( L σ µ A µ L + R σ µ A µ ) R provded QA µ QA µ Weak nterac;ons do not preserve charge conjuga;on:

Rela;on to Srednck nota;on ( g µν Dag( 1,1,1,1 ) [ N, N ] ε N [ N, N ] ε N j jk k [ N, N ] 0 j jk k j Hermtan conjugaton N N (2,1) a ε 12 ε 21 1 (c. f.g µν ) R L L T σ 2 L a ε ab b

Rela;on to Srednck nota;on ( g µν Dag( 1,1,1,1 ) [ N, N ] ε N [ N, N ] ε N j jk k [ N, N ] 0 j jk k j Hermtan conjugaton N N Conven;on: RH (do^ed) felds always (2,1) a, (1,2) [ ] a!a Dot denotes SU(2) R ndex L a R σ 2 L *!a ε!a!b!b

Rela;on to Srednck nota;on ( g µν Dag( 1,1,1,1 ) Lα a, Rα!a a χ a χ Invarants : * R χ L ( σ 2 L ) χ L T L σ 2 χ L ε ab b χ a L χ R ε!a!b!a χ b! L σ µ µ L + R σ µ µ R ξ a σ µ a!c µ ξ!c + χ!a σ µ!ac µ χ c ( ) χ σ µ µ χ + ξ σ µ µ ξ + µ ξσ µ ξ L R χ c ξ!c

Fermon Lagrangan Drac fermon: Drac equa;on Majorana fermon:

Free par;cle solu;on to the Drac equa;on

Drac- Paul bass µ γ β, βα ( ) 0 σ I 0 α β σ 0 0 I γ 0 I 0 0 σ γ 0 I σ 0 γ 5 0 I I 0 0 1 2 3 γ γ γ γ Weyl bass α σ 0 0 σ β 0 I I 0 γ 0 0 I I 0, γ 0 σ σ o, γ γ 0 γ 1 γ 2 γ 3 5 I 0 0 I { } µ ν µ ν ν µ µν γ, γ γ γ + γ γ 2g ( g µν Dag(1, 1, 1, 1) )

Free par;cle solu;on Drac Paul bass - c.f. Chapter 38 Srednck for Weyl bass solu;on ( γ µ µ m) 0 e p. x u( p) ( γ µ p µ m)u( p) ( p m)u( p) 0 µ γ ( β, βα ).e. ( α.p + βm)u Eu 0 σ I 0 α β σ 0 0 I mi σ. p ua ua E σ. p mi ub ub σ. p u ( E m) u B σ. p u ( E + m) u A A B

σ. p u ( E m) u B σ. p u ( E + m) u A A B If p 0, E + m ( E + m) u 0 u 0 B B For the 2 E>0 solu;ons, we may take ( s) ( s u ) A χ, χ 1 0 χ 0 1 (1) (2) u σ. p χ E + m ( s) ( s) B Pos;ve energy 4- spnor solu;ons of Drac s equa;on ( s) χ, 0, 1, 2 χ > E + m ( s) u N σ. p E s ( s)

For the 2 E<0 solu;ons, we may take u u σ. p σ. p χ χ E m E + m ( s) ( s ) B χ, ( s) ( s) ( s) A σ. p u ( E m) u B σ. p u ( E + m) u A A B Nega;ve energy 4- spnor solu;ons of Drac s equa;on u (s+2) N ' σ.p E + m χ (s) χ (s), E < 0, s 1,2 Orthonormal states ( r) ( s) u u 0, r s Non- rela;vs;c correspondance 1 0 0 0 2 2 2 2 (1) ( mc / ) t 0 (2) ( mc / ) t 1 (3) ( mc / ) t 0 (4) ( mc / ) t 0 e!, e!, e +!, e +! 0 0 1 0 0 0 0 1

Spn for state at rest p 0 1 0 0 0 2 2 2 2 (1) ( mc / ) t 0 (2) ( mc / ) t 1 (3) ( mc / ) t 0 (4) ( mc / ) t 0 e!, e!, e +!, e +! 0 0 1 0 0 0 0 1 Snce we have a (two- fold) degeneracy there must be some operator whch commutes wth the energy operator and whose egenvalues label the two states σ 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 3 3 Σ 3 0 σ 3 (1,2) (1,2) Egenvalues ± 1 Σ ± σ Σ 0 0 σ ( ) 2 3 2! Σ! I,! Σ has egenvalues ±! 1 1 1 2 4 2 2! Σ s spn operator S correspondng to S 1 1 2 2

Spn for state NOT at rest p 0 1 1 1 2 2 2 [ H] [ α P βm]! Σ no longer a commutng observable! Σ,! Σ,. + 0 Helcty!! 1 σ. p 0 Σ. p ",! p 2 0 σ.! p p p 0 σ I 0 α β σ 0 0 I s Egenvalues 1 +! 2 p s 1! 2 p (More generally, n arbtrary frame, spn gven by boos;ng result at rest - s µ (0, s) s ' µ Λ µ s ν ν )